Re: Does the number of nines increase?

On 7/9/2024 8:08 AM, WM wrote:

Le 09/07/2024 à 02:15, Jim Burns a écrit :

On 7/8/2024 3:57 PM, WM wrote:

Le 08/07/2024 à 19:33, Jim Burns a écrit :

A change needs be _with respect to_ something,

>
yes, to the value befor that point.

>
Yes,
for example, floor(0) with respect to floor(0-ε)
>
Or,
yes, to the value after that point
>
for example, ceiling(0) with respect to ceiling(0+ε)

>
No, my definition _for this concrete case_ is this:
A change of NUF(x) happens at point x if
for all y < x NUF(x) > NUF(y).

Then ceiling(x) _doesn't_ changeᵂᴹ "at" 0
This is a handy illustration of how your logicᵂᴹ works:
1. Decide on the answer.
2. Abuse the terms until you get that answer.

A person who believes in
loss by exchange and
NUF changing at 0

>
_near_ 0

>
So the change does not happen at zero.

In exactly the same way ceiling(x) doesn't changeᵂᴹ "at" 0

Does each nonempty set S of unit.fractionsᵂᴹ
hold a largest S.element?

>
A largest and a smallest.
Alas the smallest can only be found if it is not dark.

Each nonempty set S of unit.fractionsⁿᵒᵗᐧᵂᴹ
holds a largest S.element.
Consider the nonempty set S of unit.fractionsⁿᵒᵗᐧᵂᴹ
and the set ⤓S of unit.fractionⁿᵒᵗᐧᵂᴹ lower.bounds of S
⤓S = {u∈⅟ℕ: u≤ᴬS}
⤓S is empty or nonempty.
If ⤓S is empty,
then S doesn't hold a smallest member, which would be in ⤓S
If ⤓S is nonempty,
then unit.fractionⁿᵒᵗᐧᵂᴹ.set ⤓S holds largest member max.⤓S
If max.⤓S isn't in S
then (max.⤓S)⁺¹ is also a lower.bound of S
and in ⤓S
and larger than max.⤓S
and a contradiction.
max.⤓S is in S
max.⤓S = min.S  exists  iff
S is bounded.below in ⅟ℕ
Also,
if S is bounded.below
then each nonempty subset of S is bounded.below
and holds minimum and maximum.
Unit.fractionⁿᵒᵗᐧᵂᴹ.set S is a finiteⁿᵒᵗᐧᵂᴹ set  iff
S is bounded.below in ⅟ℕ
The set ⅟ℕ of all unit.fractionsⁿᵒᵗᐧᵂᴹ
isn't bounded below in ⅟ℕ
Each unit.fractionⁿᵒᵗᐧᵂᴹ u isn't below ⅟(1+⅟u) in ⅟ℕ
and isn't a lower.bound.
⅟ℕ is an infiniteⁿᵒᵗᐧᵂᴹ set.

Does each unit.fractionᵂᴹ u (including 1)
have a next.smaller unit.fractionᵂᴹ ⅟(1+⅟u) ?

>
Obviously not,

Then unit.fractionsᵂᴹ aren't unit.fractionsⁿᵒᵗᐧᵂᴹ
Each unit.fractionⁿᵒᵗᐧᵂᴹ (including 1) has
a next.smaller unit.fractionⁿᵒᵗᐧᵂᴹ ⅟(1+⅟u)

as I have demonstrated irrefutably

...for unit.fractionsᵂᴹ
but not for unit.fractionsⁿᵒᵗᐧᵂᴹ
This is a handy illustration of how your logicᵂᴹ works:
1. Decide on the answer.
2. Abuse the terms until you get that answer.

(refuted only by people who cannot think clear enough.
But every unit fraction that can be named
has a next smaller unit fraction.
>

Does each unit.fractionᵂᴹ v (excluding 1)
have a next.larger unit.fractionᵂᴹ ⅟(-1+⅟v) ?

>
Yes, but
for all dark unit fractions this cannot be found.
Every unit fraction excluding 1/1 that can be named
has a next larger unit fraction.

Each unit.fractionⁿᵒᵗᐧᵂᴹ (excluding 1) has
a next.larger unit.fractionⁿᵒᵗᐧᵂᴹ ⅟(-1+⅟u)

Or is  what you're talking about irrelevant to
what you're saying?

>
Relevant is this and only this:
NUF(0) = 0,
and the first step happens  at x > 0.
Like every step it is a step by 1.

If NUF(x) = 0  at  x > 0
then
⅟ℕ.greatest.lower.bound β ≥ x > 0
⅟ℕ.lower.bound ½⋅β < β
⅟ℕ.not.lower.bound 2⋅β > β
unit.fractionⁿᵒᵗᐧᵂᴹ u₂ᵦ < 2⋅β
unit.fractionⁿᵒᵗᐧᵂᴹ ¼⋅u₂ᵦ < ¼⋅2⋅β = ½⋅β
⅟ℕ.not.lower.bound ½⋅β
Contradiction.